UVa11988 Broken Keyboard(练习链表使用)

向量和数组的优势是可以随机的存取元素和在末尾添加删除元素，而当插入元素时，需要移动大量的数据，消耗大量的时间。而链表的优势是可以在O(1)删除和插入数据。所以在频繁移动元素时，可以使用链表。

分析：如果用一个数组来保存，题目中的文本随着光标位置的移动需不断的插入字符，这样会导致大量字符移动问题。解决方案是采用链表，将字符串保存在buf[1...n]中，然后用next[i]表示下标为i的字符的下一个位置的下标（链表不一定用指针实现）。为了方便起见，用一个虚拟的next[0]指向显示屏最右边的字符下标。再用cur表示光标的位置，即为光标在buf[cur]的右边.cur==0说明光标在显示屏的最左边。还用last表示显示屏最右边的字符。

#define _CRT_SECURE_NO_DEPRECATE
#include <iostream>
#include<string.h>
#include <algorithm>
using namespace std;
const int maxn = 100000 + 5;char buf[maxn];
int Next[maxn];int main(){int cur, last,n;while (scanf("%s", buf+1)!=EOF){n = strlen(buf + 1);cur = last = 0;for (int i = 1; i <= n; i++){if (buf[i] == '[')cur = 0;    //house键else if (buf[i] == ']')cur = last; //end键else{Next[i] = Next[cur];   //字符i的下一个位置为cur的下一个位置Next[cur] = i;         //curr的下一个位置为i//更新cur和lastif (cur == last)last = i;   //cur等于i表示光标在显示屏最后一个字符cur = i;  //移动光标}}Next[last] = 0;  //结束位置for (int i = Next[0]; i != 0; i = Next[i])printf("%c", buf[i]);printf("\n");}return 0;
}

转载于:https://www.cnblogs.com/td15980891505/p/5915814.html

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